Card trick thing
Kay_B
First off I wanna say that I am no mathwiz. I was wondering if anyone
could explain this card trick to me.
You need a standard deck of cards (52).
The trick is pretty much done in 2 phases.
Ace counts as 1
2 as 2
etc
knight as 11
queen as 12
king as 13
Hold the deck with the face of the cards up. Start with the first card
(it could be any card), in this example I will start with a 10. Put
the 10 down on the table, face up, then add cards ontop of the 10
until you've reached the number 13. All cards you add after the first
one will count as 1. You will now have a stack of 4 cards with the 10
as the bottom card. Turn the stack around faces down, so the 10 is on
the top.
Next card I get in my deck is a 6. I put it face up on the table and
start counting to 13. I turn the stack of 8 cards upside down (the 6
ends on top).
I then go onto the next card in the deck, and continue until I've used
the entire deck. If you have any cards left in the deck that doesnt
add up to 13, just put these cards in a separate stack, away from the
stacks you've made. These will be used later. (example: you have a 2
cards left, a 6 and a 4, max value here would be 7).
Ok, so I got 10 stacks on my table now (this is random of course). I
take away all stacks, except 3. (And add the stacks to the leftover
cards, if any).
BTW, the trick is to get someone else to lay the cards out this way,
so you couldn't know which card is the top card in each stack. Of
course you could look at the height of each stack, and estimate which
card is on top, but that is not the point.
Right, so I have 3 stacks of cards lying face down in front of me. I
reveal the top cards in 2 of the piles. In this example I got a 6 and
a 8. The idea now is to find out what the top card in the last stack
is, using the information that I have from the 2 revealed stacks,
plus the amount of cards I have in the deck (the stacks that I
removed, plus the leftovercards from earlier, if there was any).
So I start counting the cards that I have in the deck now.
I have 33 cards apart from the 3 stacks of cards. From the deck I
remove 10 cards. I now have 23 cards left in the deck (not counting
the 3 stacks on the table, of course).
As I said earlier, I had revealed 2 top cards in the 3 stacks. An 8
and a 6.
8+6 is 14.
23-14 is 9.
Which in fact is the number on the top card in the last stack.
This is probably quite chaotically written, but try it a few times and
see if you get the same result as I. If someone could explain to me
why you can always find out what the top card in the last stack is,
by this method, I would appreciate it. It's probably very simple or
very hard, cause I can't see it =)
*-----------------------*
Posted at:
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Kay_B
3 stacks with ace as top card. (3 stacksx13 cards).
You would have 13 card left in the deck. Remove 10 cards and you would
sit with 3 cards left on your hand. Ace+Ace is 2, so the last top card
would have to be an ace also, to get 3.
The other extreme. 3 stacks of kings. (3 stacks of 1 card each).
You would have 49 cards left. Remove 10 cards from the deck and you'd
have 39. You reveal 2 cards and see 2 kings. This adds up to 26.
Deduct 26 from 39 and you have 13. Last card would also have to be a
king then.
Jon Haugsand
> Hi!
>
> First off I wanna say that I am no mathwiz. I was wondering if anyone
> could explain this card trick to me.
> This is probably quite chaotically written, but try it a few times and
> see if you get the same result as I. If someone could explain to me
> why you can always find out what the top card in the last stack is,
> by this method, I would appreciate it. It's probably very simple or
> very hard, cause I can't see it =)
Quite easy. In each of the three stacks there are 14 minus "the top
card" number of cards. Say the top cards are a, b and x. Call the
remaining number of cards n, then:
52 = n + (14-a) + (14-b) + (14-x)
Which implies:
x = n-10-a-b
--
Jon Haugsand
Dept. of Informatics, Univ. of Oslo, Norway, mailto:jon...@ifi.uio.no
http://www.ifi.uio.no/~jonhaug/, Phone: +47 22 85 24 92
No Way
14 - {value top card}
The total number of cards is divided over the 4 piles, so:
52 = {leftover} + (14 - {card1}) + (14 - {card2}) + (14 - {card3})
52 = {leftover} + 42 - {card1} - {card2} - {card3}
10 = {leftover} - {card1} - {card2} - {card3}
{card3} = ({leftover} - 10) - ({card1} + {card2})
In your scenario:
{card1} = 8
{card2} = 6
{leftover} = 33
so {card3} = (33-10) - (8+6) = 23 - 14 = 9
On 6 Dec 2004 23:29:33 -0600, k...@myth-dot-as.no-spam.invalid (Kay_B)
wrote:
Kay_B
I really started to see the simplicity of the card trick after I
posted the message. Just never really thought about it. Curious
though that the number 10 solves this trick every time.
r.e.s.
-snip description of card trick-
Here's a variation with some added "mystery" ...
First give your friend a standard deck of 52 cards (facedown).
Now present a second deck (facedown), and ask your friend to
move *any* facedown cards she wishes from the second to the
first deck, counting them so you know how many she chooses to
add. Discard the smaller deck.
With the larger deck (which now has unknown cards in it) form
complete stacks as before, again with a possibly incomplete
last stack. Now ask your friend to choose any three of the
complete stacks, and combine the other stacks as 'leftover'.
Finally, have your friend reveal the top cards on any two of
the three remaining stacks. Then calculate as follows:
hidden top-card value
= (#leftover -10 -#added) -(sum of revealed top-card values).
Amazing! ;o)